|MadSci Network: Physics|
In a pendulum lab, one gets a straight line if he/she graphs the period-squared (T^2) vs. the length of the string.
The relationship equation is T = 2*pi*SQRT(length/g).
If you graph the string length on the y-axis and the T^2 on the x-axis (which would would technically be incorrect because the string length is the independant variable), the slope's units are meters/sec^2.
Does this slope represent the value of g (acceleration due to gravity)? If so, then how can it? I solved the equation T = 2*pi*SQRT(length/g) for "g."
g = length/(T/2*pi)^2
So how can one simply "exclude" the 2*pi factor and assume that the above slope represents "g." In a lab book, I see a question: "What does the slope represent? [Hint: solve your equation for length/T2.]" My answer would be that it represents nothing of importance. Only graphing "length" vs. "(T/2*pi)^2" would represent "g," which would be important.
Is my thinking off?
Re: Pendulum: What is relationship of the T^2 vs. length?
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